Does every negative semidefinite matrix lie in the convex cone generated by the squares of skew-symmetric matrices?

Question

Let $C$ be the convex cone generated by all the squares of real $n \times n$ skew-symmetric matrices.

Does every negative-semidefinite matrix lie in $C$?

I know that every square of a skew-symmetric matrix can be written in the form $$-\big(a_1 \, (x_1x_1^T + y_1y_1^T) + \cdots + a_k \, (x_kx_k^T + y_ky_k^T)\big),$$

where $a_i \ge 0$.

Answer 1

This is not true.

This can be seen easily for $n=1$ (trivially) and $n=2$ (squares of skew-symmetric matrices are multiples of the identity).

Let $n\ge2$. Then the following diagonal matrix $$ A:=\pmatrix{ -1 \\ &0\\&&\ddots\\&&&0} $$ is not in the cone. Assume otherwise. Then there are skew-symmetric matrices $A_k$ such that $A = \sum_k A_k^2$.

Multiplying this equation from the left with $e_i^T$ and from the right with $e_i$ yields $$ - \delta_{1,i} = \sum_k e_i^TA_k^2e_i = -\sum_k e_i^TA_k^TA_ke_i =-\sum_k \|A_ke_i\|^2. $$ For $i>1$ this implies that the $i$-th column of all matrices $A_k$ is zero. By skew-symmetry, the first column of all $A_k$ has to be zero as well. So $A_k=0$ for all $k$, which implies $A=0$, a contradiction.